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olasank [31]
4 years ago
14

under what conditions would a scale record a weight for an object which is equal to the magnitude of the gravitational force on

the object?
Physics
2 answers:
pentagon [3]4 years ago
7 0

answer:if there are no external forces acting on the object

Explanation:

under what conditions would a scale record a weight for an object which is equal to the magnitude of the gravitational force on the object?

If gravity and the normal reaction force are the only forces acting on the body, and the body is at rest or moving at constant velocity, the normal force and the weight are equal and opposite. Also if there are no influence of external forces

and air resistance is negligible.

from newtons law of motion which states that the rate of change in momentum is directly proportional to the force applied

f=km(v-u)/t

f=kma

where k is the constant and  equal to 1

f=ma

d1i1m1o1n [39]4 years ago
4 0
In a non accelerated frame, also called inertial frame, when no additional forces are acting upon the body.

For instance, a scale in free fall will not record the weight, but zero, like the astronauts. But also, if someone is pulling the object or pushing it, the scale will record some other value.
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Select the statements that correctly interpret the data in this heating curve.
tekilochka [14]

The consistent temperature indicated by section 2 represents the flow of energy lost as the water freezes.

Section 2 indicates that energy is being utilized to bring all molecules to the melting point.

Section 4 indicates that the temperature is stable until all molecules reach the boiling point.

The increasing temperature in section 5 indicates that molecular motion is increasing.

Explanation:

The figure is missing: find it in attachment.

The graph represents the temperature of a block of ice as heat is continuously supplied to it.

At the beginning (section 1), the temperature increases: in this phase, the heat supplied to the ice is used to increase the kinetic energy of vibration of the molecules of ice.

When the ice reach point A (melting point), the ice starts to melt. In this phase (section 2), the temperature of the substance remains constant, because all the heat supplied is used to break the bonds between the molecules, converting the ice into liquid water.

At point B, the ice has completely melted, and therefore we now have liquid water.

In sector 3 (between B and C), the temperature of the water increases again as we supply more and more heat: this heat, in fact, is used to increase the kinetic energy of the molecules, which move faster and faster.

Then, the water reaches the boiling point (point C): after that, the water starts to boil, (sector 4) and the heat supplied is entirely used to remove the intermolecular forces between the molecules of water; as a result, the temperature remains constant during the process.

At poind D, the water has completely boiled, and converted into gas state (steam).

After that, the temperature of the steam starts to increase again (sector 5) as more and more heat is supplied, because it is used to increase the kinetic energy of the molecules.

Therefore, the correct statements are:

The consistent temperature indicated by section 2 represents the flow of energy lost as the water freezes.

Section 2 indicates that energy is being utilized to bring all molecules to the melting point.

Section 4 indicates that the temperature is stable until all molecules reach the boiling point.

The increasing temperature in section 5 indicates that molecular motion is increasing.

3 0
3 years ago
Read 2 more answers
What is the pressure drop due to the bernoulli effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire
Paul [167]

Answer:

\Delta P=1581357.92\ Pa

Explanation:

Given:

  • diameter of hose pipe, D=0.09\ m
  • diameter of nozzle, d=0.03\ m
  • volume flow rate, \dot{V}=40\ L.s^{-1}=0.04\ m^3.s^{-1}

<u>Now, flow velocity in hose:</u>

v_h=\frac{\dot V}{\pi.D^2\div 4}

v_h=\frac{0.04\times 4}{\pi\times 0.09^2}

v_h=6.2876\ m.s^{-1}

<u>Now, flow velocity in nozzle:</u>

v_n=\frac{\dot V}{\pi.d^2\div 4}

v_n=\frac{0.04\times 4}{\pi\times 0.03^2}

v_n=56.5884\ m.s^{-1}

We know the Bernoulli's equation:

\frac{P_1}{\rho.g}+\frac{v_1^2}{2g}+Z_1=\frac{P_2}{\rho.g}+\frac{v_2^2}{2g}+Z_2

when the two points are at same height then the eq. becomes

\frac{P_1}{\rho.g}+\frac{v_1^2}{2g}=\frac{P_2}{\rho.g}+\frac{v_2^2}{2g}

\Delta P=\frac{\rho(v_n^2-v_h^2)}{2}

\Delta P=\frac{1000(56.5884^2-6.2876^2)}{2}

\Delta P=1581357.92\ Pa

8 0
3 years ago
Bill and Ted are standing on a bridge 40 ft above a river. Bill drops a stone, while Ted decides to throw a stone downward at 10
Y_Kistochka [10]

Answer:

D) - 0.72 secs

Explanation:

Parameters given:

Height of bridge = 40ft = 12.19 m

Initial velocity of Bill's stone = 0m/s

Initial velocity of Ted's stone = 10m/s

We find the time it take Bill's stone to bit the river and the time it takes Ted's stone to hit the river. Then we find the time difference.

Using one of the equations of motion:

For Bill:

S = ut + ½gt²

Where g = 9.8 m/s

12.19 = 0 + ½*9.8*t²

t² = 12.19/4.9 = 2.49

t = 1.58 secs

For Ted:

S = uT + ½gT²

12.19 = 10*T + ½*9.8*T²

=> 4.9T² + 10T - 12.19 = 0

Using quadratic formula and retaining only the positive value, we get that:

T = 0.86 secs

Time difference between Bill's throw and Ted's throw is:

0.86 - 1.58 = - 0.72 secs

In reality, this means that Ted must throw his stone 0.72 secs before Bill throws his for both stones to land the same time.

6 0
4 years ago
1. Define robbery, burglary, safe burglary, and theft.
neonofarm [45]

Answer:

So if a person breaks a window to get into the house or break open your cupboard to steal, it is considered burglary. Usually, an HPP covers burglary but not theft. Theft would mean the person committing the crime had access to the house or its valuables.

Please mark me the brainliest .....

8 0
2 years ago
You purchased 1.9 kg of apples from Wollaston. You noticed that they used a spring scale with the smallest division of 2.1 g to
Sphinxa [80]

Answer:

Relative Error =  0.0001%

Explanation:

What is Relative Error?

This is the ratio of the absolute error of a measurement to the measurement being taken. Relative Error is expressed as a percentage and has no units.

Given data

Weight of purchased apples= 19kg

RE = 0.0021/19

RE= 0.0001%

Hence the  Relative Error is  0.0001%

6 0
3 years ago
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